In each of the following items approximate the zeros of $\displaystyle f$ using Newton's method. Continue iterating until making two successive approximations differ at most in 0.001

The Newton iteration: $\displaystyle x_{n+1}=x_n-\displaystyle\frac{f(x_n)}{f'(x_n)}$

Well, I have a doubt about this. I'm not sure if it's asking me to iterate till $\displaystyle |x_{n+1}-x_n|\leq{0.001}|$, or if I should apply some of this:

$\displaystyle k_1>0$, $\displaystyle |f'(x)|\geq{k_1}$ and $\displaystyle |f''(x)|\leq{}k_2$ for all $\displaystyle x\in{[b]}$, then:

$\displaystyle |x_{n+1}-r|<\displaystyle\frac{k_2}{2k_1}|x_n-r|^2$

If $\displaystyle r\in{}[r-\delta,r+\delta]\subset{[a,b]}$, and $\displaystyle \delta<2(\displaystyle\frac{k_1}{k_2})$

$\displaystyle |x_{n+1}-r|<\displaystyle\frac{2k_1}{k_2}(\displaystyle\fra c{\delta}{\displaystyle\frac{2k_1}{k_2}})^2n$